If x is a positive integer, what is the remainder when 7^(12x+3) + 3 is divided by 5?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
Man Cat 4 #14-Base 7 remainder
This topic has expert replies
-
- Master | Next Rank: 500 Posts
- Posts: 125
- Joined: Mon Dec 15, 2008 9:24 pm
- dumb.doofus
- Master | Next Rank: 500 Posts
- Posts: 435
- Joined: Sat Sep 27, 2008 2:02 pm
- Location: San Jose, CA
- Thanked: 43 times
- Followed by:1 members
- GMAT Score:720
Should be 1.
It doesnt matter what the value of x is..
The way number 7 works is that the unit's digit repeats after 7^4.. so everytime you will get a sequence of 7,9,3,1.. so
I can write the equation as 7^(12x+3) = 7^12x*7^3
so 7^12x will always have units digit of 1.. when multiplied with 7^3, the unit's digit will always be 3.. now if you add 3 to the number.. the unit's digit will become 6..
So when you divide any number whose unit's digit is 6 by 5, the remainder has to be 1.
It doesnt matter what the value of x is..
The way number 7 works is that the unit's digit repeats after 7^4.. so everytime you will get a sequence of 7,9,3,1.. so
I can write the equation as 7^(12x+3) = 7^12x*7^3
so 7^12x will always have units digit of 1.. when multiplied with 7^3, the unit's digit will always be 3.. now if you add 3 to the number.. the unit's digit will become 6..
So when you divide any number whose unit's digit is 6 by 5, the remainder has to be 1.
One love, one blood, one life. You got to do what you should.
https://dreambigdreamhigh.blocked/
https://gmattoughies.blocked/
https://dreambigdreamhigh.blocked/
https://gmattoughies.blocked/
-
- Legendary Member
- Posts: 1035
- Joined: Wed Aug 27, 2008 10:56 pm
- Thanked: 104 times
- Followed by:1 members
7^(12x+3), x>0
=>7^15, 7^27, 7^39 and so on
7^1, 7^2, 7^3, 7^4 have units digits as 7, 9, 3, 1 respectively. the pattern repeats every 4th power.
7^15, 7^27, 7^39 ....all will have the units digit as 3 (as 15, 27, 39..all leave a remainder of 3 when divided by 4)
last digit of the sum of the numerator: 6. will always leave a remainder of 1 when divided by 5
hence, B
=>7^15, 7^27, 7^39 and so on
7^1, 7^2, 7^3, 7^4 have units digits as 7, 9, 3, 1 respectively. the pattern repeats every 4th power.
7^15, 7^27, 7^39 ....all will have the units digit as 3 (as 15, 27, 39..all leave a remainder of 3 when divided by 4)
last digit of the sum of the numerator: 6. will always leave a remainder of 1 when divided by 5
hence, B
-
- Master | Next Rank: 500 Posts
- Posts: 125
- Joined: Mon Dec 15, 2008 9:24 pm